A projectile is shot from the ground at a velocity of 4 m/s at an angle of pi/6. How long will it take for the projectile to land?

1 Answer
Joaquín M.
Mar 3, 2016

4* sqrt(3)/5 m

Explanation:

Let's start with Newton's equation **F** = m* **a** . As the problem is bi-dimensional, the previous equation can be split in the horizontal and vertical components

F_x = m*a_x for the horizontal component and

F_y = m*a_y for the vertical component.

The horizontal acceleration is null while the vertical acceleration is equal to gravity

a_x = 0 ; a_y =-g

and using the definition of acceleration

a_x = (d^2x)/dt^2 ; a_y = (d^2y)/dt^2

and integrating both part of equations

x= v_(x0)*t ; y= v_(yo)*t - g*t^2/2 (1)

The projectile will land when y=0 => introducing it in (1)

0= v_(yo)*t_l - g*t_l^2/2 ; t_l = 2*v_(yo)/g

and replacing in the equation or x

x_l= 2*v_(x0)*v_(y0)/g

The components of shot speed are

v_(x0)= v_0 *cos(pi/6)

v_(y0)= v_0 *sin(pi/6)

thus

x_l= 2*v_0^2/g*cos(pi/6)*sin(pi/6)

and replacing by the figures given in the problem

x_l= 2*16/10 m*cos(pi/6)*sin(pi/6) = 16/5m*1/2*sqrt(3)/2 = =4* sqrt(3)/5 m

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